Question

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Answer 1

$\text{Step 1 : Formula of Energy stored in capacitor}$

The Energy stored $U$ in a capacitor of Capacitance $C$, Charge $Q$ and Potential Difference $V$ across it is given by:

$U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}=\frac{1}{2}QV$ $....\left(1\right)$

$\text{Step 2: Calculation of Energy stored}$

Given, Capacitance $C=12pF=12\times {10}^{-12}F$

And Potential difference $V=50V$

So using the first formula from Equation $\left(1\right)$, we get Energy stored as:

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times [12\times {10}^{-12}F]\times \left[\right(50V{)}^2]=15\times {10}^{-9}J$

$U=15nJ$

Hence Energy stored in capacitor is $15nJ$.